The radius of the base of a cylinder is increasing at a rate of $1$ meter per hour and the height of the cylinder is decreasing at a rate of $4$ meters per hour. At a certain instant, the base radius is $5$ meters and the height is $8$ meters. What is the rate of change of the volume of the cylinder at that instant (in cubic meters per hour)? Choose 1 answer: Choose 1 answer: (Choice A) A $-180\pi$ (Choice B) B $180\pi$ (Choice C) C $20\pi$ (Choice D) D $-20\pi$ The volume of a cylinder with base radius $r$ and height $h$ is $\pi r^2h$.
Solution: Setting up the math Let... $r(t)$ denote the base radius of the cylinder at time $t$, $h(t)$ denote the height of the cylinder at time $t$, and $V(t)$ denote the volume of the cylinder at time $t$. We are given that $r'(t)=1$ and $h'(t)=-4$ (notice that $h'$ is negative). We are also given that that $r(t_0)=5$ and $h(t_0)=8$ for a specific time $t_0$. We want to find $V'(t_0)$. Relating the measures The measures relate to each other through the formula for the volume of a cylinder: $V(t)=\pi[r(t)]^2h(t)$ We can differentiate both sides to find an expression for $V'(t)$ : $V'(t)=2\pi r(t)r'(t)h(t)+\pi[r(t)]^2h'(t)$ Using the information to solve Let's plug ${r(t_0)}={5}$, ${r'(t_0)}={1}$, ${h(t_0)}={8}$, and $C{h'(t_0)}=C{-4}$ into the expression for $V'(t_0)$ : $\begin{aligned} V'(t_0)&=2\pi{r(t_0)}{r'(t_0)}{h(t_0)}+\pi[{r(t_0)}]^2C{h'(t_0)} \\\\ &=2\pi({5})({1})({8})+\pi({5})^2(C{-4}) \\\\ &=-20\pi \end{aligned}$ In conclusion, the rate of change of the volume of the cylinder at that instant is $-20\pi$ cubic meters per hour. Since the rate of change is negative, we know that the volume is decreasing.